11th Class Chemistry Redox Reactions

  • question_answer 10) Permanganate (VII) ion, \[MnO_{4}^{-}\] in basic solution oxidises iodide ion, \[{{I}^{-}}\] to produce molecular iodine \[({{I}_{2}})\] and manganese (IV) oxide\[(Mn{{O}_{2}})\]. Write the balanced ionic equation to represent this redox reaction.


    Reaction: \[MnO_{4}^{-}(aq)+{{I}^{-}}(aq)\to Mn{{O}_{2}}(s)+{{I}_{2}}(s)\] 1st step: Splitting into two half reactions: (balance the elements other than hydrogen and oxygen) \[MnO_{4}^{-}\to Mn{{O}_{2}}(s)\] (Reduction) \[2{{I}^{-}}\to {{I}_{2}}(s)\] (Oxidation) 2nd step: Balance oxygen atoms by adding water molecule to the oxygen deficient side. \[MnO_{4}^{-}\to Mn{{O}_{2}}+2{{H}_{2}}O\] \[2{{I}^{-}}\to {{I}_{2}}\] 3rd step: Balance hydrogen by adding \[{{H}^{+}}\] ions to the hydrogen deficient side. \[MnO_{4}^{-}+4{{H}^{+}}\to Mn{{O}_{2}}+2{{H}_{2}}O\] \[2{{I}^{-}}\to {{I}_{2}}\] 4th step: Balance the charge by adding electrons \[MnO_{4}^{-}+4{{H}^{+}}+3{{e}^{-}}\to Mn{{O}_{2}}+2{{H}_{2}}O\] \[2{{I}^{-}}\to {{I}_{2}}+2{{e}^{-}}\] 5th step: Equations are added in such a way that electrons are cancelled; we now get the balanced equation. \[[MnO_{4}^{-}]+4{{H}^{+}}+3{{e}^{-}}\to Mn{{O}_{2}}+2{{H}_{2}}O]\times 2\] \[[2{{I}^{-}}\to {{I}_{2}}+2{{e}^{-}}]\times 3\] \[\frac{\frac{{}}{2MnO_{4}^{-}+6{{I}^{-}}+8{{H}^{+}}\to 2Mn{{O}_{2}}+3{{I}_{2}}+4{{H}_{2}}O}}{Balanced\text{ }equation}\] \[2MnO_{4}^{-}+4{{H}_{2}}O\to 2Mn{{O}_{2}}+3\underset{(Basic\,medium)}{\mathop{{{I}_{2}}+8O{{H}^{-}}}}\,\]  

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